At least an orthogonal projection inside a side Let $P$ a point inside a convex n-agon and let $P_1, P_2, ..., P_n$ the ortogonal projections of $P$ on the sides of the n-agon. How can I show that at least one of these projections lies inside a side of the poligon? I tried to prove that a convex n-agon is dibisible in $n$ triangle with $P$ and two vertex of the n-agone as vertex with all the angles $\leq 90°$ but I failed... any advice/solution? Thanks :)

Assume that's not the case, and each orthogonal projection is outside the polygon.

Let $A$ be the closest polygon vertex to $P$, and let $B$ be a neighboring vertex. Let's find the locus points such that $P_1 = \text{proj}_{AB}(P) \
otin AB$.

There are two possibilities.

1. $\angle PAB > \frac{\pi}{2}$. Construct a line $g$ perpendicular to $PA$ through $A$. Then $B$ must be in the half-plane bordered by $g$ not containing $X$.
2. $\angle PBA > \frac{\pi}{2}$. In this case B is inside the circle with diameter PA. However, since $A$ was chosen to be the vertex closest to $P$, this is impossible.



Finally, $A$ has two neighbors. However, having both of then in the half-plane described below line $g$ violates convexity, a contradiction.

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